What is the integral from-2 to x of (10t^2-9t) ?

The integral from-2 to x of (10t^2-9t) is (10(x^3+8))/3-9((x^2)/2-2)

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integral from-2 to x of (10t^2-9t)

\int_{- 2}^{x} (10 t^{2} - 9 t) d t

\frac{10 ( x ^{3} + 8 )}{3} - 9 (\frac{x ^{2}}{2} - 2)

Solution steps

\int_{- 2}^{x} 10 t^{2} - 9 t d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 2}^{x} 10 t^{2} d t - \int_{- 2}^{x} 9 t d t

\int_{- 2}^{x} 10 t^{2} d t = \frac{10 ( x ^{3} + 8 )}{3}

\int_{- 2}^{x} 9 t d t = 9 (\frac{x ^{2}}{2} - 2)

= \frac{10 ( x ^{3} + 8 )}{3} - 9 (\frac{x ^{2}}{2} - 2)

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What is the integral from-2 to x of (10t^2-9t) ?
The integral from-2 to x of (10t^2-9t) is (10(x^3+8))/3-9((x^2)/2-2)